Answer:
50m
Explanation:
Given parameters:
Initial velocity = 20m/s
Acceleration = 4m/s²
Time = 10s
Unknown:
Distance traveled by the rocket = ?
Solution:
To solve this problem use the expression below;
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
final velocity = 0
Insert the parameters and solve;
0² = 20² + 2 x 4 x s
-400 = 8s
s = 50m
Disregard the negative sign because distance cannot be negative.
At the player's maximum height, their velocity is 0. Recall that

which tells us the player's initial velocity
is

The player's height at time
is given by

so we find their airtime to be

Answer:
a. 299,792,458 m/s
Explanation:
Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.
Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
